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Math Help - [SOLVED] Equation to plane through 3 pts

  1. #1
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    [SOLVED] Equation to plane through 3 pts

    I have points p, q, r,

    p = (-4,-1,-1)
    q = (-2,0,1)
    r = (-1,-2,-3)

    Okay so I need to find the equation to the plane using these three points.


    Step 1. I find vector pq and vector qr

    vector pq = (2,1,2)
    vector qr = (1,-2,-4)

    Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)

    Step 3. I sub this into any point using (I forget what it's called form) and I get

    0(x+4) + 6(y+1) + -5(z+1) = 0
    6y - 5z +1 = 0

    I think that's right, what do you think?
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  2. #2
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    Quote Originally Posted by thekrown View Post
    I have points p, q, r,

    p = (-4,-1,-1)
    q = (-2,0,1)
    r = (-1,-2,-3)

    Okay so I need to find the equation to the plane using these three points.


    Step 1. I find vector pq and vector qr

    vector pq = (2,1,2)
    vector qr = (1,-2,-4)

    Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)

    Step 3. I sub this into any point using (I forget what it's called form) and I get

    0(x+4) + 6(y+1) + -5(z+1) = 0
    6y - 5z +1 = 0

    I think that's right, what do you think?
    Neither q nor r are on that plane...

    Tonio
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  3. #3
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    What am I doing wrong? Are my steps correct or are my calculations wrong?
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  4. #4
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    Quote Originally Posted by thekrown View Post
    What am I doing wrong? Are my steps correct or are my calculations wrong?
    \overrightarrow {PQ}  = \left\langle {2,1,2} \right\rangle \,\& \,\overrightarrow {PR}  = \left\langle {3, - 1, - 2} \right\rangle
    \overrightarrow {PQ}  \times \,\overrightarrow {PR}  = \left\langle {0,2, - 1} \right\rangle
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  5. #5
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    Quote Originally Posted by thekrown View Post
    I have points p, q, r,

    p = (-4,-1,-1)
    q = (-2,0,1)
    r = (-1,-2,-3)

    Okay so I need to find the equation to the plane using these three points.


    Step 1. I find vector pq and vector qr

    vector pq = (2,1,2)
    vector qr = (1,-2,-4)

    Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)
    pq x qr= (0, 10, -5)

    Step 3. I sub this into any point using (I forget what it's called form) and I get

    0(x+4) + 6(y+1) + -5(z+1) = 0
    6y - 5z +1 = 0

    I think that's right, what do you think?
    You should at least have put p, q, and r into that equation to see for yourself!
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